(a) Use vector components to find the magnitude and direction of F. (b) Check the reasonableness of your answer in part (a) by doing a graphical solution approximately to scale.
SOLUTION
The figure below shows how the $xy$-plane is overlapped on the system (upper left). A diagram in right corner shows how it looks in an xy-plane where the x-axis is horizontal and y-axis is vertical.
Since the forces are balanced then the resultant vector, $\vec{R}=0$, then
$\vec{R}=0=\vec{F}+\vec{W}+ \vec{pull}$. Re-arranging to obtain $\vec{F}$,
$\vec{F}=-\vec{W}- \vec{pull}$
The components of the given vectors where the angles are measured from the +x-axis rotating towards the +y-axis are$W_x=(124 \ N) \ cos \ (270^o-40.0^o)=-79.71 \ N$
$W_y=(124 \ N) \ sin \ (270^o-40.0^o)=-94.99 \ N$
$pull_x=(100 \ N) \ cos \ (30.0^o)=86.60 \ N$
$pull_y=(100 \ N) \ sin \ (30.0^o)=50.0 \ N$
Then the components of $\vec{F}$ are
$F_x=-W_x-pull_x=-(-79.71 \ N)-86.60 \ N=-6.89 \ N$
$F_y=-W_y-pull_y=-(-94.99 \ N)-50.0 \ N=44.99 \ N$
solving for the magnitude and direction using the Pythagorean theorem and trigonometry,
$F=\sqrt{(F_x)^2+(F_y)^2}=45.5 \ N$
$\theta=arctan \ (\frac{F_y}{F_x})=-81.3^o$ measured clockwise from the -x-axis.
